Partial fractions

[renegade05]

I am studying for my CALC II final and just brushing up on my integration using the partial fractions technique.

I remember learning this topic and something that i never understood was the repeated linear factors. My text and teacher never showed how this works, just how to do it. I HATE WHEN THEY DO THIS.

Anyway, here is my confusion:

Say you have:

\(\displaystyle \[ \int \frac{x}{(x+3)^2}\,dx\]\)

I know to solve this you must set it up like this:

\(\displaystyle \int\frac{x}{(x+3)^2}\,dx=\)\(\displaystyle \int \frac{A}{(x+3)}+\frac{B}{(x+3)^2}\,dx\)

I just wanna know why do we split it like this? Why does the A term get an \(\displaystyle (x+3)\) and the B term gets an \(\displaystyle (x+3)^2\) ? I mean i can do it no problem just due to memorizing the technique, but i would just like to know how this works.

Am I just reading into this too much ? Am I missing something easy here?

Thanks!

 

[galactus]

This is due to a theorem in advanced algebra which says that every rational function of the form \(\displaystyle \frac{P(x)}{Q(x)}\) in which the degree of the

numerator is less than the degree of the denominator can be written as the sum:

\(\displaystyle \frac{P(x)}{Q(x)}=F_{1}(x)+F_{2}(x)+\cdot\cdot\cdot F_{n}(x)\)

where the F's are rational functions of the form:

\(\displaystyle \frac{A}{(ax+b)^{k}}, \;\ \frac{Ax+B}{(ax^{2}+bx+c)^{k}}\)

The terms on the right side, the F's, are called partial fractions and the entire right side is the decomposition.

If you do enough algebra, you can figure out how to do these without partial fractions.

Take your example:

\(\displaystyle \frac{x}{(x+3)^{2}}\)

Add and subtract x+3:

\(\displaystyle \frac{x-(x+3)+(x+3)}{(x+3)^{2}}\)

\(\displaystyle \frac{x-x-3+(x+3)}{(x+3)^{2}}\)

\(\displaystyle \frac{-3}{(x+3)^{2}}+\frac{1}{x+3}\)

Done

 

[soroban]

Hello, renegade05!

When we have a repeated linear factor in the denominator,
. . we must have a separate fraction "for each power".

\(\displaystyle \text{For example: }\:\frac{1}{(x-1)\underbrace{(x-2)^3}} \:=\:\frac{A}{x-1} + \underbrace{\frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3}}\)


\(\displaystyle \text{Solution: }\:A = \text{-}1,\;B = 1,\;C = \text{-}1,\;D = 1\)

\(\displaystyle \text{Therefore: }\:\frac{1}{(x-1)(x-2)^3} \;=\;-\frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{(x-2)^2} + \frac{1}{(x-2)^3}\)